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I finished a program to do connected component analysis using union - find algorithm. Is it true that the complexity of the code is \$O(N logN)\$, where \$N\$ is the total number of pixels (512x512 by example, not 512).

Is there any way to improve the performance?

import cv2
import numpy as np
import random

class QuickUnionUF:

    def __init__(self, N):
        self.id = range(N)
        self.sz = [0] * N

    @classmethod
    def fromimage(self, im):
        self.id = im
        self.sz = [0] * len(im)

    def root(self, i):
        while (i != self.id[i]):
            i = self.id[i]
        return i

    def getresult(self):
        result = [self.root(i) for i in self.id]
        return result

    def connected(self, p, q):
        return self.root(p) == self.root(q)

    def union(self, p, q):
        i = self.root(p)
        j = self.root(q)

        if (i == j):
            return
        if (self.sz[i] < self.sz[j]):
            self.id[i] = j
            self.sz[j] += self.sz[i]
        else:
            self.id[j] = i
            self.sz[j] += self.sz[i]        

def bwlabel(im):

    M, N = im.shape[:2]
    qf = QuickUnionUF(M * N)
    for i in range(M - 1):
        for j in range(N - 1):
            if (im[i][j] == im[i][j+1]):
                qf.union(i * N + j, i * N + j + 1)
            if (im[i + 1][j] == im[i][j]):
                qf.union(i * N + j, (i + 1) * N + j)

    mask = np.reshape(np.array(qf.getresult()), (M, N))
    values = np.unique(mask).tolist()

    random.seed()
    colors = [(random.randint(0,255), random.randint(0,255), random.randint(0,255)) for k in range(len(values))]

    out = np.zeros((M, N, 3))
    for i in range(M):
        for j in range(N):
            label = values.index(mask[i][j])
            out[i,j] = colors[label]

    return out

im = cv2.imread("bw.jpg",cv2.IMREAD_GRAYSCALE)
out = bwlabel(im > 100)
cv2.imwrite("result.png", out)
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Is it true that the complexity of the code is O(NlogN), where N is the total number of pixels (512x512 by example, not 512).

You main loop in bwlabelis:

for i in range(M - 1):
    for j in range(N - 1):
        if (im[i][j] == im[i][j+1]):
            qf.union(i * N + j, i * N + j + 1)
        if (im[i + 1][j] == im[i][j]):
            qf.union(i * N + j, (i + 1) * N + j)

So you are calling qf.union \$O(N*M) = O(n)\$ times. There you call self.root twice:

def root(self, i):
    while (i != self.id[i]):
        i = self.id[i]
    return i

What's the worst case running time of this? At most \$O(n)\$, since self.id has \$n\$ values and there are no cycles. However, since you are comparing ranks in union:

    if (self.sz[i] < self.sz[j]):
        self.id[i] = j
        self.sz[j] += self.sz[i]
    else:
        self.id[j] = i
        self.sz[j] += self.sz[i]        

You will have \$O(log\ n)\$ paths to root, so the total running time is \$O(n\ log\ n)\$.

Is there any way to improve the performance?

Yes, path compression. You descended down two paths from p and q to i and j and it took \$O(log\ n)\$. At each step you can point a node to its root, instead of just the next element. You can do this recursively like so:

def root(self, i):
    if i == self.id[i]:
        return i
    self.id[i] = self.root(self.id[i])
    return self.id[i]

Recursion adds its own cost, though, so whether it's faster in practice would require testing. An alternative would be to add a second iterative compression function:

def compress(self, i, r):
    while i != r:
        i, self.id[i] = self.id[i], r

Now you are walking the path twice, which again may or may not be faster than recursion. You should only need to call this for one of the paths, though.

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